High School Evaluating a Complex Expression with Given Value

[anemone]

Evaluate $\large(1+x^5-x^7)^{{2014}^{1007}}$ if $x=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{20}}$.

--------------------
Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solutions::)

1. Rido12
2. laura123
3. Olok
4. lfdahl
5. Pranav
6. kaliprasad

Solution from Rido12:

Spoiler

First, evaluate $x$:

$\begin{align*}x&=\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{20}}+\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{20}}\\&=\sqrt{\dfrac{30+10\sqrt{5}}{100}}+\sqrt{\dfrac{30-10\sqrt{5}}{100}}\\&=\sqrt{\dfrac{25+5+10\sqrt{5}}{100}}+\sqrt{\dfrac{25+5-10\sqrt{5}}{100}}\\&=\frac{5+\sqrt{5}}{10}+\frac{5-\sqrt{5}}{10}\\&=1\end{align*}$Therefore,

$$\large(1+x^5-x^7)^{{2014}^{1007}}=\large(1)^{{2014}^{1007}}=1$$

Solution from Pranav;

Spoiler

$$x=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{20}} \Rightarrow x^2=\frac{6+2\sqrt{5}+6-2\sqrt{5}+2\sqrt{36-10}}{20}=1$$
Hence,

$$\left(1-x^5+x^7\right)^{2014^{1007}}=\left(1-x^5\left(1-x^2\right)\right)^{2014^{1007}}=\boxed{1}$$